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zloy xaker [14]
3 years ago
6

The opening to a cave is a tall, 30-cm-wide crack. A bat that is preparing to leave the cave emits a 30 kHz ultrasonic chirp. Ho

w wide is the "sound beam" 100 m outside the cave opening? Use vsound = 340 m/s.
Physics
1 answer:
sweet-ann [11.9K]3 years ago
5 0

Answer:

Width of sound beam is 7.557 m

Explanation:

First we will calculate the wave length from given data:

λ=v/f

Were:

v is the speed

f is the frequency

\lambda=\frac{340}{30*10^3}\\ \lambda=0.01133 m

We considered the opening long and narrow, Using single slit diffraction formula:

mλ=dsinΘ

where:

d is the crack width

m is the order

Θ is angle

Considering m=1, The angle between first minimum from center of beam is:

\theta=sin^{-1}(\frac{m\lambda}{d})\\\theta=sin^{-1}(\frac{1*0.01133}{30*10^{-2}})\\ \theta=2.164^o

The width of beam is:

tanΘ=y/L

tan\theta=\frac{w/2}{L}\\ w=2L\ tan\theta\\w=2*100*tan 2.164\\w=7.557 m

Width=7.557 m

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The time taken for the tiny saliva to travel is 0.55 second.

The  horizontal distance traveled at speed of 4 m/s is 2.2 m.

The horizontal distance traveled at speed of 20 m/s is 11 m.

<h3>Time of motion of the tiny saliva</h3>

The time taken for the tiny saliva to travel is calculated as follows;

h = vt + ¹/₂gt²

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